Question

A thin uniform rod OA of length L, mass M is pivoted to ground at end O in a vertical plane. The rod is free to rotate about a horizontal axis through O. It is released from rest in position OA. When it is in position OB, the angular speed of rod is

(1) \([\frac{(3\sqrt{3}-1)g}{2L}]^{\frac{1}{2}}\)

(2) \([\frac{(3\sqrt{3}-1)g}{L}]^{\frac{1}{2}}\)

(3) \([\frac{(\sqrt{3}-1)g}{2L}]^{\frac{1}{2}}\)

(4) \([\frac{(3\sqrt{3}+1)g}{2L}]^{\frac{1}{2}}\)

Answer 1

 (1) \([\frac{(3\sqrt{3}-1)g}{2L}]^{\frac{1}{2}}\)

The loss is gravitational P.E in moving from position OA to position OB.

This energy is converted into rotational K.E. of rod. Let ω be angular speed of rod in position OB.

Gain in rotational K.E 

From law of conservation of energy